Question: The equation of a circle $C$ is $x^2+y^2+12x+8y+16 = 0$. What is its center $(h, k)$ and its radius $r$ ?
To find the equation in standard form, complete the square. $(x^2+12x) + (y^2+8y) = -16$ $(x^2+12x+36) + (y^2+8y+16) = -16 + 36 + 16$ $(x+6)^{2} + (y+4)^{2} = 36 = 6^2$ Thus, $(h, k) = (-6, -4)$ and $r = 6$.